Question

我正在尝试启动我的Hibernate项目，但因为我的Customer类中的一个属性字段是Password类型，项目将无法启动，因为它无法创建一个Db列。

我尝试使用属性字段上方的@Column(columnDefinition="varchar(20)")注释密码字段，并且在getter上方但都没有帮助。

有人提出任何线索吗？

这是班级：

package be.mobyus.ie5.entities;


import sun.security.util.Password;

import javax.persistence.*;
import java.util.Collection;

@Entity
public class Customer {

    @Id
private Long id;
private String username;

private Password password;
private String name;
private String firstName;

@OneToMany(targetEntity=Order.class, fetch= FetchType.EAGER)
private Collection<Order> orders;

/***********************************************/

    public Customer(){

    }

    /***********************************************/


    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    public String getUsername() {
        return username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

    @Column(columnDefinition = "character varying(22)")
    public Password getPassword() {
        return password;
    }

    public void setPassword(Password password) {
        this.password = password;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public String getFirstName() {
        return firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }

    public Collection<Order> getOrders() {
        return orders;
    }

    public void setOrders(Collection<Order> orders) {
        this.orders = orders;
    }
}

我收到以下错误：

Exception in thread "main" org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'sessionFactory' defined in class path resource [applicationContext-hibernate.xml]: Invocation of init method failed; nested exception is org.hibernate.MappingException: Could not determine type for: sun.security.util.Password, at table: Customer, for columns: [org.hibernate.mapping.Column(password)]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1628)
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.doCreateBean(AbstractAutowireCapableBeanFactory.java:555)
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.createBean(AbstractAutowireCapableBeanFactory.java:483)
    at org.springframework.beans.factory.support.AbstractBeanFactory$1.getObject(AbstractBeanFactory.java:306)
    at org.springframework.beans.factory.support.DefaultSingletonBeanRegistry.getSingleton(DefaultSingletonBeanRegistry.java:230)
    at org.springframework.beans.factory.support.AbstractBeanFactory.doGetBean(AbstractBeanFactory.java:302)
    at org.springframework.beans.factory.support.AbstractBeanFactory.getBean(AbstractBeanFactory.java:197)
    at org.springframework.beans.factory.support.DefaultListableBeanFactory.preInstantiateSingletons(DefaultListableBeanFactory.java:742)
    at org.springframework.context.support.AbstractApplicationContext.finishBeanFactoryInitialization(AbstractApplicationContext.java:866)
    at org.springframework.context.support.AbstractApplicationContext.refresh(AbstractApplicationContext.java:542)
    at org.springframework.context.support.ClassPathXmlApplicationContext.<init>(ClassPathXmlApplicationContext.java:139)
    at org.springframework.context.support.ClassPathXmlApplicationContext.<init>(ClassPathXmlApplicationContext.java:83)
    at be.mobyus.ie5.SpringRunner.main(SpringRunner.java:10)
Caused by: org.hibernate.MappingException: Could not determine type for: sun.security.util.Password, at table: Customer, for columns: [org.hibernate.mapping.Column(password)]
    at org.hibernate.mapping.SimpleValue.getType(SimpleValue.java:455)
    at org.hibernate.mapping.SimpleValue.isValid(SimpleValue.java:422)
    at org.hibernate.mapping.Property.isValid(Property.java:226)
    at org.hibernate.mapping.PersistentClass.validate(PersistentClass.java:597)
    at org.hibernate.mapping.RootClass.validate(RootClass.java:265)
    at org.hibernate.boot.internal.MetadataImpl.validate(MetadataImpl.java:329)
    at org.hibernate.boot.internal.SessionFactoryBuilderImpl.build(SessionFactoryBuilderImpl.java:451)
    at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:710)
    at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:726)
    at org.springframework.orm.hibernate5.LocalSessionFactoryBean.buildSessionFactory(LocalSessionFactoryBean.java:511)
    at org.springframework.orm.hibernate5.LocalSessionFactoryBean.afterPropertiesSet(LocalSessionFactoryBean.java:495)
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFactory.java:1687)
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1624)
    ... 12 more

Answer 1

您可以使用spring框架中的Base64Util来加密/解密Password属性并将其作为varchar存储在DB中

Hibernate 5.2 Spring 4.3 MappingException：无法确定：sun.security.util.Password的类型

1 个答案: